Question

The filament of a light bulb has a temperature of 3.38 x 103 °C and radiates 60.0 W of power. The emissivity of the filament is 0.437. Find the surface area of the filament.

Answer 1

Answer on Question #37945, Physics, Other

One might find the surface area of the light bulb from Stefan-Boltzmann law:

$$P = S \in \sigma T ^ {4} \quad ,$$

where $P$ is the power radiated, $\epsilon$ is the emissivity, $\sigma$ is Stefan's constant ($\sigma = 5.67 \cdot 10^{-8} J \cdot s^{-1} \cdot m^{-2} \cdot K^{-4}$) and $T$ is temperature.

Surface are is hence

$$S = \frac {P}{\epsilon \sigma T ^ {4}} = \frac {6 0 W}{0 . 4 3 7 \cdot 5 . 6 7 \cdot 1 0 ^ {- 8} J \cdot s ^ {- 1} \cdot m ^ {- 2} \cdot K ^ {- 4} \cdot (3 . 3 8 \cdot 1 0 ^ {3} + 2 7 3 . 1 5) ^ {4} K ^ {4}} = 1 3. 6 \cdot 1 0 ^ {- 6} m ^ {2}.$$